Problems & Puzzles:
Puzzles
Puzzle 7. Palindrome  Primes: 2
questions:
 Patrick De Geest, observes that:
10499 + 10501 + 10513 = 31513
is the first case of a prime
palindrome (31513) obtained adding three consecutive
primes, the central one of them being a palindrome prime
also (10501), and asks if this will ever happen again?
( http://www.worldofnumbers.com/palprim5.htm
)
 By my own I ask if exists any
palindrome prime obtained adding the first k
primes, that is to say:
What is the first k, such that:
Nk = 2 + 3 + 5 + 7 + 11 +…..+
pk = Prime Palindrome?
Solution
Question b.
Jud McCranie sends his results on
this problem :
"One of your puzzles asks for
the sum of the first k primes to be a palindromic prime. Except
for the trivial solutions k=1, sum=2 and k=2, sum=5;
there aren't any more for k <= 54,400,028 (Pk =
1073741989, sum Nk = 28422918403819825)" (9/July/98)
***
Vasiliy Danilov sent us (Fri. 17 Jul.
1998) the following palindromes (notprimes) of his
search for this puzzle :
k
Pk
Sum k
8
19
77
7693
78347 285080582
8510
87641 352888253
12941
139241 854848458
146134 1959253
137372273731
637571 9564097
2939156519392
Pk < 100000000 and program stopped.
He also sent us some
references at the Neil Sloane’s records (A038582 ,
A038584, A038583,) partially related with this puzzle.
***
Judson McCranie adds this
two lines (Jul/23/98) of palindromic notprimes, as a
result of his search ... "I've now checked k through
105097565"
k
Pk Sum k (palindromic)
27198825
516916921 6833383883833386
53205635
1048924213 27155268786255172
So, it seems to me that the
answer to this puzzle is "negative"...
***
Question a.
Almost five years later, Giovanni Resta came up with several (8) solutions, as the asked by Patrick
I have found some solutions to puzzle 7a (that is to
find more examples like
10499 + _10501_ + 10513 = _31513_
where there are 3 consecutive primes, the middle one
palindromic, whose sum is palindromic prime. Here they are (at least they
are easier to check !)
324456535654403 + 324456535654423 + 324456535654553 =
973369606963379
326151616151597 + 326151616151623 + 326151616151659 =
978454848454879
332526262625197 + 332526262625233 + 332526262625369 =
997578787875799
11708180908180709 + 11708180908180711 +
11708180908180733 =
35124542724542153
12829181718192793 + 12829181718192821 +
12829181718192869 =
38487545154578483
12907072927070909 + 12907072927070921 +
12907072927070953 =
38721218781212783
12918271817281913 + 12918271817281921 +
12918271817281949 =
38754815451845783
32928090809082919 + 32928090809082923 +
32928090809082947 =
98784272427248789
***
One week later J. K. Andersen
came up with the following big solutions to question a), gotten with a
clever approach (as usual):
Question a
I have found many big solutions with a pattern.
Let p be a palindrome where the first/last digit
and all digits in even positions are in {0,1,2} and all other digits
(odd internal positions) are in {7,8,9}. Then 3p+20 is also a
palindrome. Some of Resta's solutions follow this pattern.
My strategy: Generate such palindromes p and 3p+20. If both are prp
then find the prp right before and after p. If they sum to
3p+20
then we have a probable solution and prove the 4 primes.
First a C program using Michael Scott's
Miracl bigint library found the solutions with prp's. PrimeForm/GW
doublechecked that the 3 prp's are consecutive.
A 73digit solution with middle palindromic
prime:
p =
1070707070707070707070707071829182829282819281707070707070707070707070701
The solution equation is (p52) + (p) + (p+72) =
(3p+20)
=
3212121212121212121212121215487548487848457845121212121212121212121212123
The 4 primes were proved with Tony Forbes'
VFYPR.
A 303digit solution:
p = 10^302 + 10*(7*10^30170)/99 +
10^140*22001012200100221010022 + 1 Let w be "0" followed by 69
concatenations of "70". Then p is the concatenation
1w92708082907170928080729w1 Solution equation: (p138) + (p) + (p+158)
= (3p+20) These 4 primes were proved with Marcel Martin's Primo.
After trying 25098 prp's as the middle
palindrome I got this 527digit
solution:
p = 10^526 + 10*(7*10^52570)/99 +
10^251*2102210022021202200122012 + 1 Let w be "0" followed by 124
concatenations of "70". Then p is the concatenation
1w728092807290919092708290827w1 The equation is (p180) + (p) +
(p+200) = (3p+20), proved with Primo.
The complexity of my algorithm is around
O(digits^5). I will not search a titanic solution. It is about 25
times harder than 527 digits and would probably take weeks.
***
